Problem: $f\,^{\prime}(x)=-\dfrac{4}{x^2}$ and $f(2)=4$. $f(1) = $
Solution: Finding $f(x)$ We have $f'(x)=-\dfrac{4}{x^2}$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (-\dfrac{4}{x^2})\,dx \\\\ & = {4x^{-1}} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(2)=4$. Here's what we get when we plug in $2$ : $\begin{aligned}f(2)&={4(2)^{-1}} {+ C}\\\\ &={2} {+ C} \end{aligned}$ We are given that this must equal $4$ : $4 = {2} {+ C}$ Solving the equation gives us ${C=2}$. Finding $f(2)$ Now, we have that $f(x)= {4x^{-1}} {+2}$. Let's find $f(1)$ by plugging in $1$ : $\begin{aligned}f(1)&=4(1)^{-1}+2\\\\ &=6 \end{aligned}$ The answer $f(1) = 6$